# Problems with Electric Field questions

## Homework Statement

Find the point on the x-axis where the net electric field is zero for two particles of charges q1=1x10^-9C and q2=2x10^-9C. Assume q1 and q2 are 20cm apart. Assume q1 is located at x=0.

I solved this but it's not the same as the book's answer.. GAH! What do you guys get? Thanks.

## Homework Equations

Enet=E1+E2, Enet=0 => |E1|=|E2|

E=kq/r² where q is the charge creating the field

## The Attempt at a Solution

E1=kq1/r²
E2=kq2/(.2m - r)²

kq1/r²=kq2/(.2m - r)²
q1/r²=q2/(.2m - r)²
q1(.2m - r)²=q2r²
q1(.2m - r)²=q2r²
q1(0.4m-0.4mr+r²)-q2r²=0 <------ There's my mistake
(q1)0.4m-(q1)0.4mr+(q1)r²-q2r²=0
(q1-q2)r²-(q1)0.4mr+(q1)0.4m=0
((1E-9C)-(2E-9C))r²-(1E-9C)0.4mr+(1E-9C)0.4m=0
(-1E-9C)r²-(1E-9C)0.4mr+(1E-9C)0.4m=0
(-1E-9C)r²-(4E-10C)mr+(4E-10C)m=0
(-1E-9)r²-(4E-10)r+(4E-10)=0
r=-0.863325 or r=0.463325

The answer in the book is 8.3cm. Where did I go wrong?
------
Edit:
q1(.2m - r)²=q2r²
q1(0.04m-0.4mr+r²)-q2r²=0
(q1)0.04m-(q1)0.4mr+(q1)r²-q2r²=0
(q1-q2)r²-(q1)0.4mr+(q1)0.04m=0
((1E-9C)-(2E-9C))r²-(1E-9C)0.4mr+(1E-9C)0.04m=0
(-1E-9C)r²-(1E-9C)0.4mr+(1E-9C)0.04m=0
(-1E-9C)r²-(4E-10C)mr+(4E-11C)m=0
(-1E-9)r²-(4E-10)r+(4E-11)=0
Positive r comes out to 8.3cm

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Related Introductory Physics Homework Help News on Phys.org
do you realise that Q2 has charge two times of Q1? Simply suibstitute Q2 with 2Q1 and cancel out the Q1s and your equations will appear much simpler.

I managed to simplify the equation into a quadratic equation offhand. You should arrive at a quadratic equation.

Last edited:
Right, you can see:

(-1E-9)r²-(4E-10)r+(4E-10)=0 before I found my mistake and
(-1E-9)r²-(4E-10)r+(4E-11)=0 after I found my mistake.